A158065 a(n) = 36*n + 1.

37, 73, 109, 145, 181, 217, 253, 289, 325, 361, 397, 433, 469, 505, 541, 577, 613, 649, 685, 721, 757, 793, 829, 865, 901, 937, 973, 1009, 1045, 1081, 1117, 1153, 1189, 1225, 1261, 1297, 1333, 1369, 1405, 1441, 1477, 1513, 1549, 1585, 1621

Offset: 1

Author

Vincenzo Librandi, Mar 12 2009

Keywords

• nonn
• easy

Comments

The identity (36*n + 1)^2 - (36*n^2 + 2*n)*6^2 = 1 can be written as a(n)^2 - A158064(n)*6^2 = 1. - Vincenzo Librandi, Feb 11 2012

Parametrize Pythagorean triangles with parameters a and b and side lengths x = b^2 - a^2, y = 2*a*b and z = a^2 + b^2. Generate one Pythagorean triangle with a=n-1 and b=n and side lengths (x1, y1, z1), and another one with a=n, b=n+1 and side lengths (x2, y2, z2). Then 2*a(n) = x2 - x1 + 12*(y2-y1) + 6*(z2-z1). - J. M. Bergot, Jul 16 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(6^2*t+2)).
• John Elias, Illustration of Initial Terms: Hexagram of Triangular Perimeters
• Index entries for linear recurrences with constant coefficients, signature (2, -1).

Crossrefs

Cf. A008585, A158064.

Programs

Magma

I:=[37, 73]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 11 2012

Mathematica

Range[37, 2000, 36] (* Vladimir Joseph Stephan Orlovsky, Jun 15 2011 *)
LinearRecurrence[{2, -1}, {37, 73}, 50] (* Vincenzo Librandi, Feb 11 2012 *)

PARI

for(n=1, 40, print1(36*n + 1", ")); \\ Vincenzo Librandi, Feb 11 2012

Formula

G.f.: x*(37-x)/(1-x)^2. - Vincenzo Librandi, Feb 11 2012

a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 11 2012

a(n) = 12*A008585(n) + 1 (see illustration in links). - John Elias, Jun 29 2021